Differential EQNS: $y \, dx = \left[ x + (y^2 - x^2)^{1/2} \right] dy$

Sydney Sales's picture

ydx = x + √ y^2 - x^2 dy

Jhun Vert's picture

Please use proper groupings by using parenthesis. Hindi kasi malinaw ang problem.

Sydney Sales's picture

ydx= x + ( y^2 - x^2) ^ 1/2 dy

Jhun Vert's picture

The x don't have differential element, do you mean this?
$y \, dx = x + (y^2 - x^2)^{1/2} \, dy$

Or this?
$y \, dx = \left[x + (y^2 - x^2)^{1/2} \right] dy$

Sydney Sales's picture

ydx = [x + ( y^2 - x^2 )^ 1/2] dy

this one.

Jhun Vert's picture

Solution
$y \, dx = \left[x + (y^2 - x^2)^{1/2} \right] dy$

Let x = vy
dx = v dy + y dv

$y(v\,dy + y\,dv) = \left[vy + (y^2 - v^2y^2)^{1/2} \right] dy$

$vy\,dy + y^2\,dv = \left[vy + y(1 - v^2)^{1/2} \right] dy$

$y^2\,dv = y(1 - v^2)^{1/2}\,dy$

$y\,dv = (1 - v^2)^{1/2}\,dy$

$\dfrac{dv}{(1 - v^2)^{1/2}} = \dfrac{dy}{y}$

$\displaystyle \int \dfrac{dv}{\sqrt{1 - v^2}} = \int \dfrac{dy}{y}$

$\arcsin v = \ln y + c$

$\arcsin \left( \dfrac{x}{y} \right) = \ln y + c$

Sydney Sales's picture

sa factoring po. bakit po naging.

[ vy + y (1- v^2 )^1/2 ] dy

Jhun Vert's picture

(y2 - v2 y2)1/2
  = [ y2 (1 - v2) ] 1/2
  = y2(1/2) (1 - v2)1/2
  = y (1 - v2)1/2

Sydney Sales's picture

salamat po.

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